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2 §9.1 Basic ConceptsAttractor: a state toward which the system evolves intime starting from certain initial conditions.Basin of attraction: the set of initial conditions whichinitiates the evolution terminating in the attractor.Fixed point: if an attractor is in a form of a uniquepoint in state space.Limit Cycle: if an attractor consists of a periodicsequence of states.

5 The complete matrix description of the linear portion of the system shown in the figure is given bynet = WV + i - twherenetiT111netTi2i =t =2net =2netiTnnnare vectors containing activation, external inputto each neuron and threshold vector, respectively.

7 §9.2 Discrete-Time Hopfield NetworkAssuming that the neuron’s activation function is sgn,the transition rule of the i-th neuron would be-1, if net < 0 (inhibited state)v(*)+1, if net > 0 (excitatory state)If, for a given time, only a single neuron is allowed toupdate its output and only one entry in vector v isallowed to change, this is an asynchronous operation,under which each element of the output vector isupdated separately while taking into account the mostrecent values for the elements that have already beenupdated and remain stable.

8 Based on (*), the update rule of a discrete-time recurrent network, for one value of i at a time, becomesK+1tkv = sgn(w v + i - T ) for random i, i=1, 2, …, nand k=0, 1, 2, ...iiiiwhere k denotes the index of recursive update. This isreferred as asynchronous stochastic recursion of theHopfield network. This update process will continueuntil all n entries of v have been updated.The recursive computation continues until the outputnode vector remains unchanged with further iterations.Similarly, for synchronous operation, we haveK+1iK+1v = T[Wv + i - t], for all neurons, k=0, 1, ...kwhere all neurons change their output simultaneously.

9 Geometrical Explanation The output vector v is one of the vertices of the n-dimensional cube [-1, 1] in E space. The vectormoves during recursions from vertex to vertex, untilit is stabilizes in one of the 2 vertices available.The movement is from a vertex to an adjacent vertexsince the asynchronous update mode allows for asingle-component update of an n-tuple vector at atime.The final position of v as k, is determined byweights, thresholds, inputs, and the initial vector vas well as the order of transitions.nnn

10 To evaluate the stability property of the dynamical system of interest, the computational energy functionis defined in n-dimensional output space v .If the increments of a certain bounded positive-valuedcomputational energy function under the transitionrule are found to be non-positive, then the functioncan be called a Lyapunov function, and the systemwould be asymptotically stable.The scalar-valued energy function for the discussedsystem is a quadratic form:ntttE = - 1/2 V WV - i V + t V

13 Further we can show that the non-increasing energy function has a minimum.Since W is indefinite because of its zero diagonal, thenE has neither a minimum nor maximum inunconstrained output space. However, E is obviouslybounded in n-dimensional space consisting of the 2vertices of n-dimensional cube, Thus, E has to reachits minimum finally under the update algorithm.nExample of recursive asynchronous update ofcomputed digit 4:

14 (b)(a)(c)(d)(e)where (a) k=0, (b) k=1, (c) k=2, (d) k=3, (e) k=4.The initial map is a destroyed digit 4 with 20% of thepixels randomly reversed. For k>4, no changes areproduced at the network output since the systemarrived at one of its stable states.

15 §9.3 Gradient-Type Hopfield Network Consider the continuous-time single-layer feedbacknetworks. One of its model is given below.iiiiwn2311nw32wn1ggggu123n2uun1u3cccc123nvvvv123nvvvv23n1

17 It follows thus that the change of E, in time, are in the Then we havedu(t)C= Wv(t) - Gu(t) + i andv(t) = f(u(t))dtIt can be shown thatdEdutdv= - (c )< 0dtdtdtIt follows thus that the change of E, in time, are in thegeneral direction toward lower values of he energyfunction in v space -- the stability condition.n

18 §9.4 Feedback Networks for Computational ApplicationsIn principle, any optimization problems whose objectivefunction can be expressed in the form of energy functioncan be solved by feedback networks convergence.Take the Traveling Salesman Problem as an example.Given a set of n cities A, B, C, … with pairwise distancesd , d , … try to find a close tour which visits each cityonce, returns to the starting city and has a minimumtotal path length.ABACThis is an NP-complete problem.

19 To map this problem onto the computational network, we require a representation scheme, in which the finallocation of any individual city is specified by the outputstates of a set of n neurons.E.g., n=5, the neuronal state (5 neurons) shown belowwould represent a tour:2OrderCityABCDE

20 In the nn square representation, this means that in an output state describing a valid tour there can be onlyone “1” in each row and each column, all other entriesbeing “0”.2In this scheme, the n symbols v will be described bydouble indicies, v : x stands for city name, j for theposition of that city in tour.ixjTo enable the N neurons to compute a solution to theproblem, the network must be described by an energyfunction in which the lowest energy state correspondsto the best path of the tour.

21 An appropriate form for this function can be found by considering the high gain limit, in which all finalnormal output will be 0 or 1. The space over which theenergy function is minimized in this limit is the 2corners of the N-dimensional hypercube defined byv =0 or 1.NiConsider those corners of this space which are the localminima stable states) of the energy functionBCA2   v v +   v v +(  v - n)E =222xixjxiyixi1xijiixyxxiwhere A, B, C are positive constants, v  {0, 1}.i-- The first term is 0 iff each city row x contains nomore than one “1”.

22 -- The second term is 0 iff each position in tour column i contains no more than one “1”.-- The third term is 0 iff there are n cities of “1” in theentire matrix.Thus, this energy function evaluated on the domain ofthe corners of the hypercube has minima with E = 0for all stable matrices with one “1” in each row andcolumn. All other states have higher energy.1Hence, including these terms in an energy functiondescribing a TSP network strongly favors stable stateswhich are at least valid tour in the TSP problem.

23 Another requirement, that E favors valid tours representing shout path, is fulfilled by adding oneadditional term to E . This term contains informationabout the length of the path corresponding to a giventour, and its form can be1DE =   d v (v v )22xyxiy,i+1y,i-1xyxiwhere subscripts are defined modulo n, in order toexpress easily “end effects” such as the fact that then-th city on a tour is adjacent in the tour to both city(n-1) and city 1, i.e., v = v . Within the domain ofstates which characterizes a valid tour, E is numeric-ally equal to the length of the path for that tour.y,n+jy,j2

24 If A, B, and C are sufficiently large, all the really low energy states of a network described by this functionwill have the form of a valid tour. The total energy ofthat state will be the length of the tour, and the stateswith the shortest path will be the lowest energy states.Using the row/column neuron labeling schemedescribed above for each of the two indicies, theimplicitly defined connection matrix isT = - Ad (1-d ) “inhibitory connections within each row”- Bd (1-d )“inhibitory connections with each column”- C “global inhibition”- D (d d ) “data term”xi,yjxyijijxyxyj,i+1j,i-1The external input are I = 2C “excitation bias”nxi

25 The “data term” contribution, with D, to T is the input which describes which TSP problem (I.e., wherethe cities actually are) is to be solved.xi,yjThe term with A, B, C provide the general constraintsrequired for any TSP problem.The “data term” contribution controls which one of then! Set of these properly constrained final states isactually chosen as the “best” path.The problem formulated as shown below has beensolved numerically for the continuous activationfunction with l =50, A=B=D=250, and C=100, for10  n 30. Quite satisfactory solution has been found.

27 By properly selecting the weights, it is possible to §9.5 Associative MemoryBy properly selecting the weights, it is possible tomake the stable states of the network just be the ones,M, we want to store.Under this condition, the network’s state should notchange if the network is initially in the state M;whereas if not in M, it is expected that the network’sstable state should be the ones, in M, closest to theinitial state (in the sense of Hamming distance).There are two categories of AM:1) Auto-AM: If input x’=x +v, where x {x , …, x },then output y=x .aa1Maax +vax

28 2) Hetero-AM:If x’=x +v, where x y , …, x y stored then output y=y .One of the tasks is how to find the suitable weightssuch that the network perform a function of AM?The most frequently used rule for this purpose is theOuter Product Rule. Assume that-- Consider an n-neuron network;-- Each activity state x {-1, 1};-- Hebbean rule is observed: w = x x , > 0.iaaijij